Competition problem — Geometry — Olympiad Maths Prep

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Geometry Difficulty 5.1 AIME, harder Prove it Romania

Let $O$ be the circumcircle of a triangle $ABC$ . A circle $k$ is tangent to the lines $BC$ , $CA$ , $AB$ at points $D, E, F$ , respectively, such that $A$ is on the other side of the line $BC$ with respect to the circle. Suppose the circle $k$ is equal to the circumcircle of the triangle. Prove that lines $OD$ and $EF$ are perpendicular.

Solution

Let $I_a$ be the center of the circle $k$ and let $T$ be the midpoint of the arc $BC$ - not containing $A$ - of the circumcircle $ABC$ . Notice that $OT$ is the perpendicular bisector of the line segment $BC$ to deduce that $OT \perp BC$ . As $I_a D \perp BC$ and $OT = I_a D = R$ , the quadrangle $ODI_aT$ is a parallelogram. Consequently $OD \parallel TI_a$ , or, equivalently, $OD \parallel AI_a$ .

On the other hand, since $AE = AF$ and $AI_a$ bisects angle $\angle FAE$ , the lines $AI_a$ and $EF$ are perpendicular, and so are lines $OD$ and $EF$ .

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